sum_{k=1}^{infty} frac{1}{k !} left(sum_{n=1} | vedclass

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(D) The given sum is $S = \sum_{k=1}^{\infty} \frac{1}{k !} \left(\sum_{n=1}^k 2^{n-1}\right)$.Inside the bracket,we have a geometric series with $a=1

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$e^2-e$

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(D) The given sum is $S = \sum_{k=1}^{\infty} \frac{1}{k !} \left(\sum_{n=1}^k 2^{n-1}\right)$.Inside the bracket,we have a geometric series with $a=1$,$r=2$,and $k$ terms,so $\sum_{n=1}^k 2^{n-1} = \frac{1(2^k-1)}{2-1} = 2^k-1$.Substituting this into the expression,we get $S = \sum_{k=1}^{\infty} \frac{2^k-1}{k !}$.This can be split into two sums: $S = \sum_{k=1}^{\infty} \frac{2^k}{k !} - \sum_{k=1}^{\infty} \frac{1}{k !}$.We know that $e^x = \sum_{k=0}^{\infty} \frac{x^k}{k !} = 1 + \sum_{k=1}^{\infty} \frac{x^k}{k !}$,so $\sum_{k=1}^{\infty} \frac{x^k}{k !} = e^x - 1$.For the first sum,$x=2$,so $\sum_{k=1}^{\infty} \frac{2^k}{k !} = e^2 - 1$.For the second sum,$x=1$,so $\sum_{k=1}^{\infty} \frac{1^k}{k !} = e^1 - 1 = e - 1$.Thus,$S = (e^2 - 1) - (e - 1) = e^2 - e$.

$\sum_{k=1}^{\infty} \frac{1}{k !} \left(\sum_{n=1}^k 2^{n-1}\right)$ is equal to

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